मराठी

If logα8 = γ, logβα = –1 and log1/4β = –1 then αβγ(1α+1)log5(β2+4γ2) is equal to ______.

Advertisements
Advertisements

प्रश्न

If logα8 = γ, logβα = –1 and log1/4β = –1 then `(1/α + 1)^(log_sqrt(5)(β^2 + 4γ^2)` is equal to ______.

पर्याय

  • `sqrt(5)`

  • 5

  • 25

  • 625

MCQ
रिकाम्या जागा भरा
Advertisements

उत्तर

If logα8 = γ, logβα = –1 and log1/4β = –1 then `(1/α + 1)^(log_sqrt(5)(β^2 + 4γ^2)` is equal to 625.

Explanation:

logα8 = γ, logβα = –1, log1/4β = –1

⇒ β = 4, α = `1/4`, γ = `-3/2`

⇒ `(1/α + 1)^(log_sqrt(5)(β^2 + 4γ^2)) = 5^(log_sqrt(5)(16 + 9)`

= `5^(log_(5 1/2) 5^2)`

= 54

= 625

shaalaa.com
Fundamental Integrals Involving Logarithms Functions
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×